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When a metallic surface is illuminated with light of wavelength $\lambda $, the stopping potential is $X$ volt. When the same surface is illuminated by the light of wavelength $2\lambda $, stopping potential is $\dfrac{X}{3}$. Threshold wavelength for the metallic surface is:
A) $\dfrac{{4\lambda }}{3}$
B) $4\lambda $
C) $6\lambda $
D) $\dfrac{{8\lambda }}{3}$

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Last updated date: 22nd Jun 2024
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Answer
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Hint: When the light of a certain wavelength strikes on a metal surface, electrons get emitted from the metal surface. This electron is known as photoelectron and the phenomenon of emission of electrons from the metal surface, when exposed to the light, is known as the photoelectric effect. The prefix ‘photo’ is a Greek word that means light.

Complete step by step solution:
We know that Einstein’s photoelectric equation is given by,
$\dfrac{{e{V_o}}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$ …….. (1)
Where ${V_o}$ is the stopping voltage, $\lambda $ is the corresponding wavelength of the light and ${{\lambda _o}}$ is the threshold wavelength of the light.
For a wavelength $\lambda $ the corresponding stopping voltage is X. Hence putting, ${V_o} = X$ in equation (1) we get,
$\dfrac{{eX}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$ ……… (2)
For a wavelength ${2\lambda }$ the corresponding stopping voltage is X. Hence putting, ${V_o} = \dfrac{X}{3}$ and replace ${\lambda }$ with ${2\lambda }$ in equation (1) we get,
$\dfrac{{eX}}{{3hc}} = \dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}$ …….. (3)
Dividing equation (2) by equation (3) we get,
$ \Rightarrow 3 = \dfrac{{\dfrac{1}{\lambda } - a}}{\begin{array}{l}\dfrac{1}{{2\lambda }} - a\\\end{array}}$
Where $a = \dfrac{1}{{{\lambda _o}}}$ for the sake of simplification of calculations.
 $ \Rightarrow 3 = \dfrac{{1 - \lambda a}}{{\dfrac{{1 - 2a\lambda }}{2}}}$
$ \Rightarrow 3 - 6a\lambda = 2 - 2a\lambda $
$ \Rightarrow a = \dfrac{1}{{4\lambda }}$
$ \Rightarrow \dfrac{1}{{{\lambda _o}}} = \dfrac{1}{{4\lambda }}$
$\therefore {\lambda _o} = 4\lambda $

The value of threshold wavelength is found to be $4\lambda $ hence we can conclude that option B is the correct answer option.

Note: Let us understand the difference between threshold frequency and threshold wavelength.
1. A threshold frequency is the minimum frequency of incident radiation which causes the photoelectric effect to occur. Below the threshold frequency, the photoelectric effect does not occur. The threshold frequency is denoted by ${\nu _o}$. It is measured in Hz.
2. A threshold wavelength is a maximum wavelength of incident radiations which causes the photoelectric effect to occur. Above the threshold wavelength, the photoelectric effect does not occur. The threshold frequency is denoted by ${\lambda _o}$.
The threshold frequency and threshold wavelength have an inverse relationship.